A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9928 m at Tokyo and 0.9943 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
(g Cambridge / g Tokyo)?
any help?
T=2PI(L/g)
so the ratio is ...
1=sqrt(.9928/gtokio*gengland/.9943)
squaring both sides..
gengland/gtokyo= .9943/.9928
awesome, Thanks a lot...
To find the ratio of the free-fall accelerations at the two locations, you can use the formula for the period of a pendulum:
T = 2π√(L/g)
Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
Given that the period of the pendulum is 2.000 s, the length of the pendulum in Tokyo is 0.9928 m, and the length of the pendulum in Cambridge is 0.9943 m, we can set up the following equations:
2.000 = 2π√(0.9928/gTokyo) --> Equation 1
2.000 = 2π√(0.9943/gCambridge) --> Equation 2
We want to find the ratio gCambridge/gTokyo. To do this, we can divide Equation 2 by Equation 1:
2π√(0.9943/gCambridge) / 2π√(0.9928/gTokyo)
Canceling out the common terms, we get:
√(0.9943/gCambridge) / √(0.9928/gTokyo)
Simplifying further, we get:
√(0.9943/0.9928) * √(gTokyo/gCambridge)
The term √(0.9943/0.9928) is approximately equal to 1.0015, so the ratio simplifies to:
1.0015 * √(gTokyo/gCambridge)
Therefore, the ratio of the free-fall accelerations at Cambridge to Tokyo is approximately 1.0015 multiplied by the square root of the ratio of gTokyo to gCambridge (gCambridge / gTokyo).
To find the ratio of the free-fall accelerations at Tokyo and Cambridge, we need to use the equation for the period of a pendulum. The period of a pendulum is given by the formula:
T = 2π √(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
We are given the period to be precisely 2.000 s for both locations. Let's solve for g for each location using the given lengths of the seconds pendulum.
For Tokyo:
Given length of the seconds pendulum (L) = 0.9928 m
Substituting these values into the equation, we have:
2 = 2π √(0.9928/g_Tokyo)
Now, let's solve for g_Tokyo.
Dividing both sides by 2π gives:
1 = √(0.9928/g_Tokyo)
Squaring both sides gives:
1 = 0.9928/g_Tokyo
Rearranging the equation to solve for g_Tokyo, we have:
g_Tokyo = 0.9928/1
g_Tokyo = 0.9928 m/s²
For Cambridge:
Given length of the seconds pendulum (L) = 0.9943 m
Using the same steps as above, we can find g_Cambridge:
2 = 2π √(0.9943/g_Cambridge)
1 = √(0.9943/g_Cambridge)
1 = 0.9943/g_Cambridge
g_Cambridge = 0.9943/1
g_Cambridge = 0.9943 m/s²
Finally, we can calculate the ratio of the free-fall accelerations at the two locations:
(g_Cambridge / g_Tokyo) = (0.9943 m/s²) / (0.9928 m/s²)
(g_Cambridge / g_Tokyo) ≈ 1.00151
Therefore, the ratio of the free-fall accelerations at Cambridge to Tokyo is approximately 1.00151.